Sun 13 January 2019

How information can update Conditional Probability

Written by Hongjinn Park in Articles

Here's a problem that I found very interesting.

Why would it matter that a girl was born in March?

A couple has two kids and the kids are equally likely to be boy or girl. The gender of the first one is independent of the second.

$$P(\text{girl,girl}) = \frac{P(g,g)}{P(g,g)+P(g,b)+P(b,g)+P(b,b)} = \frac{1}{4}$$ $$P(\text{girl,girl} \mid \text{first kid is a girl}) = \frac{P(g,g)}{P(g,g)+P(g,b)} = \frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{4}} = \frac{1}{2} = P(\text{second kid is a girl})$$ $$P(\text{girl,girl} \mid \text{at least one is a girl}) = \frac{P(g,g)}{P(g,g)+P(g,b)+P(b,g)} = \frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}} = \frac{1}{3}$$ $$P(\text{girl,girl} \mid \text{at least one is a March born girl}) = \frac{P(\text{both girls, at least one is a March born girl})}{P(\text{at least one is a March born girl})}$$

Then using multiplication rule on the numerator,

$$= \frac{P(\text{both girls})[1-P(\text{both not March born girls} \mid \text{both girls})]}{1-P(\text{both not March born girls})} =\frac{(1/4)(1-(11/12)^2)}{1-(23/24)^2} = \frac{23}{47} \approx 0.489$$

Punch line: Note that

$$P(\text{girl,girl} \mid \text{at least one is a girl}) = 0.333 < 0.489 = P(\text{girl,girl} \mid \text{at least one is a March born girl}) < 0.5$$

Why is this the case? Because when given information that helps identify a specific individual, the answer gets closer to $1/2$. For example, "first kid is a girl" clearly identifies a specific one. Another example, assuming all $365$ days equally likely and day of births are independent,

$$P(\text{girl,girl} \mid \text{at least one is a girl born on 4/20}) =\frac{(1/4)(1-(364/365)^2)}{1-(729/730)^2} \approx 0.4997$$


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